Question: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $k \neq 0$. $x = \dfrac{k}{3(4k - 3)} \div \dfrac{6k}{40k - 30} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $x = \dfrac{k}{3(4k - 3)} \times \dfrac{40k - 30}{6k} $ When multiplying fractions, we multiply the numerators and the denominators. $x = \dfrac{ k \times (40k - 30) } { 3(4k - 3) \times 6k } $ $ x = \dfrac {k \times 10(4k - 3)} {6k \times 3(4k - 3)} $ $ x = \dfrac{10k(4k - 3)}{18k(4k - 3)} $ We can cancel the $4k - 3$ so long as $4k - 3 \neq 0$ Therefore $k \neq \dfrac{3}{4}$ $x = \dfrac{10k \cancel{(4k - 3})}{18k \cancel{(4k - 3)}} = \dfrac{10k}{18k} = \dfrac{5}{9} $